Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sturm-Liouville question

  1. Jun 7, 2006 #1
    View attachment 7093
    I have a question that pertains to the Sturm-Liouville theory. Prove that {sin((pi)nx/a)} n=1 (a>0) is the basis for L2 (0,a).
    Last edited: Jan 20, 2007
  2. jcsd
  3. Jun 12, 2006 #2


    User Avatar
    Homework Helper

    A basis set for L^2 (0,a)

    I will assume we are dealing with continuous, real-valued functions here, as in your posted theorem. A set of functions, say [tex]\left\{ f_{n}(x)\right\}_{n=1}^{\infty},[/tex] forms a basis for L2(0,a) if and only if

    for [tex]g(x)\in L^2(0,a)[/tex] we have [tex]\left< f_{n}(x),g(x)\right> =0[/tex] for every [tex]n=1,2,\ldots[/tex] implies that [tex]g(x)=0,[/tex]

    where <,> denotes the inner product on [tex]L^2(0,a)[/tex] defined by [tex]\left< g(x) , h(x)\right> = \int_{0}^{a}g(x)h(x) dx[/tex]

    In the above requirement for a set of functions to be basis, it should be understood that

    [tex]\left< f_{n}(x),g(x)\right> =0[/tex] for every [tex]n=0,1,2,\ldots[/tex]

    means that

    [tex]\left< f_{0}(x),g(x)\right> =\left< f_{1}(x),g(x)\right> =\cdots =0[/tex]

    or equivalently that g(x) is orthogonal to every fn(x),

    the "implies that [tex]g(x)=0,[/tex]" part means that and the only function g(x) for which this condition may be satisfied (if the set of functions be a basis) is the zero function.

    Now since we are dealing with continuous, real-valued functions, the Stone-Weierstrass theorem (in particular, Weierstrass approximation theorem) applies and any [tex]g(x)\in L^2(0,a)[/tex] can be approximated by a polynomial with arbitrary precision, hence we will test our set of functions against [tex]g(x)=c_0+c_1x+c_2x^2+\cdots +c_mx^m=\sum_{k=0}^{m}c_kx^k[/tex] for [tex]m=0,1,2,\ldots[/tex], where not every ck is zero. Here it goes:

    [tex]\left< f_{n}(x) , g(x)\right> = \int_{0}^{a}\left( \sum_{k=0}^{m}c_kx^k \right) \sin\left( \frac{\pi nx}{a} \right) dx = \sum_{k=0}^{m}c_k\int_{0}^{a}x^k \sin\left( \frac{\pi nx}{a} \right) dx[/tex]

    then prove that if the last integral is zero for every n=1,2,..., then [tex]c_k=0[/tex] for k=0,1,2,...m and you're done. You might try using [tex]\sin (x) = \frac{e^{ix}-e^{-ix}}{2i}[/tex] to evaluate the integral.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook